LeetCode Perfect Squares
原题链接在这里：https://leetcode.com/problems/perfect-squares/
题目：
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
题解：
Let dp[i] denotes the least number of perfect squares sum to i.
Then for all the candiates smaller than i, if the difference between i and candidate is perfect square, then update the dp[candidate]+1. Maintain the smallest.
Thne how to make sure the difference is perfect square.
Let candidate = i - j*j.
Time Complexity: O(nlogn).
Space: O(n).
AC Java:
class Solution {
public int numSquares(int n) {
int [] dp = new int[n+1];
for(int i = 1; i<=n; i++){
dp[i] = i;
for(int j = 0; i-j*j>=0; j++){
dp[i] = Math.min(dp[i], dp[i-j*j]+1);
}
}
return dp[n];
}
}
Could also do it from head to tail.
初始化i*i的位置为1, 然后对i + j*j 更新min(dp[i+j*j], dp[i] + 1).
Time Complexity: O(nlogn). Space: O(n).
AC Java:
public class Solution {
public int numSquares(int n) {
if(n < 0){
return 0;
}
int [] dp = new int[n+1];
Arrays.fill(dp, Integer.MAX_VALUE);
for(int i = 0; i*i <= n; i++){
dp[i*i] = 1;
}
for(int i = 1; i<=n; i++){
for(int j = 1; i+j*j<=n; j++){
dp[i+j*j] = Math.min(dp[i+j*j], dp[i]+1);
}
}
return dp[n];
}
}
与Count Primes类似.
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